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Sagot :
Using the hypergeometric distribution, it is found that the probability is 12/132
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The rolls are chosen from a sample without replacement, which means that the hypergeometric distribution is used to solve this question.
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Hypergeometric distribution:
The probability of x successes is given by the following formula:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]
In which:
x is the number of successes.
N is the size of the population.
n is the size of the sample.
k is the total number of desired outcomes.
Combinations formula:
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
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- 12 rolls means that [tex]N = 12[/tex]
- 4 defective means that [tex]k = 4[/tex]
- 2 are chosen, which means that [tex]n = 2[/tex]
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What is the probability of selecting two defective rolls?
This is P(X = 2). So
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]P(X = 2) = h(2,12,2,4) = \frac{C_{4,2}*C_{8,0}}{C_{12,2}} = \frac{4 \times 3}{12 \times 11} = \frac{12}{132}[/tex]
12/132 is the probability.
A similar problem is found at https://brainly.com/question/24229905
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